(1) when a=9 and b=8, 3ab41× 6 = 239046;

(2) when a=0 and b= 1, 3ab41× 6 =180846;

So 180846 ≤ Y 2 ≤ 239046.

Then 425 < y < 489

Because the unit number of the product of 3ab4 1×6 must be 6, the unit number of y can be 4 or 6, and y 2 must be a multiple of 6, so y must also be a multiple of 6;

So between 425 and 489, the matching y

There may be: y=426, 444, 456, 474, 486.

Then the corresponding: (y 2)/6 = 30246, 32856, 34656, 37446, 39366.

Because the product of 3ab4 1×6 must have 4 decimal places, it is qualified: (y 2)/6 = 30246 or 37446;

From this, we can get a=0 (at present, 0 is a natural number in primary school mathematics) and b = 2；; Or a=7 and b=4.